中国石油大学北京机械04级机械原理试卷答案

PAGE1PAGE7ChinaUniversityPetroleum,Beijing,2006-2007,Semester1Testpaper(A)ofMechanismsandmachineTheory.(Closedbook)No.1234567TotalScoreScoreClass           Name         StudentIDNo.             1.Inacrank-slidermechanismshowninthefigure,thecrankisthedrivinglink.thestrokeHoftheslideris50mm,thecoefficientoftraveloftravelspeedvariationK=1.5,theoffsete=20mm.Pleasecalculate:(20%)1)Thecrankacuteanglebetweenthetwolimitingpositions，2)Thedimensionsofcranka,couplerb，3)ThemaximumpressureangleMAX.ofthemechanism，4)Willthedead-pointoccurifthesliderisdrivinglink?评分标准：本题每小题５分图画对二极限位置之夹角可得3分，写出公式k=(180º+θ)/(180º-θ)可得４分。结果1分计算方法可不同，必要公式和中间过程4分，结果１分写对表达式即可得５分，位置对，但结果错误扣１分。答对得分，错不得分。2.Drawthekinematicdiagramofthemechanism.(8%)评分标准：每错一处运动副扣２分，未按比例画图不扣分3.Inthecombinedgeartrainshownbelow,Z1=17,Z2=68,Z3=23,Z4=19.Z4’=20,Z5=24.Findthetrainratioi15.(10%)Solution:Intheordinarygeartrain1-2：①Intheplanetarygeartrain3-4=4’-5-2:②③评分标准：公式①４分，公式中丢“—”扣1分，公式②5分结果1分。4.Thefig.shownisaninvolutespurgearset:thedrivinggear1rotatesatangularvelocityω1,ra1andra2aretheradiiofbothaddendumcircleofgear1andgear2respectively,rb1andrb2aretheradiiofbothbasecircleofgear1andgear2.modulem＝4mm，pressureangleα＝20°，μl＝0.001m/mm.Drawinthefigureandpointoutthefollowing:(20%)1)TheactuallineofactionB1B2graphically,2)ThetheoreticallineofactionN1N2graphically,3)MeasuretheactuallineofactionB1B2,,Calculatethecontactratioεαaccordingthedefinitionofcontactratioεα=B1B2/Pbpb=mcos评分标准：找对B1B2即可得６分，否则不得分。找对N1N2可得6分，N1N2方向反扣２分量得B1B2可得2分，写出εα的定义式得６分5.Calculatethedegreeoffreedomofthemechanismshown(theformulashouldbegiven).Addtheserialnumberstothelinksandindicatecompoundhinge,passiveDOF,redundantconstraintbeforethecalculationoftheDOF.Whatistheconditionforthemechanismstohavedeterminedmotion?(10%)评分标准n=8１分pL=11１分pH=1１分F=3n－2pL－pH=3×8-2×11-1４分＝1１分机构有确定运动的条件为：原动件数与机构的自由度数相等。１分指示对compoundhinge,passiveDOF,redundantconstraint共１分6.AmechanicalsystemrunsataperiodicfluctuatedspeedwithaangularperiodofφT=2π,arotatinglinkisselectedasanequivalentlink.TheequivalentdrivenmomentMrontheequivalentlinkaregivenfunctionsoftherotatingangleasshowninthefigure.andtheequivalentinputmomentMdontheequivalentlinkisconstant.TheequivalentmomentofinertiaJecanbeneglected.Theaveragespeedisnm=1000r/min.Theallowablecoefficientofspeedfluctuationis[]=0.01（12%）1)FindtheequivalentinputmomentMd,2)CalculatethemaximumincrementofworkWmax,3)CalculatetheminimummomentofinertiaoftheflywheelJF.SupposeE0=E0Ea=E0-F1=E0+0.5×0.25×30=E0+15/4Eb=Ea+F2=E0－15/4Ec=Eb-F3=E0+15/4Ed=Ec+F4=E0－15/4Ef=E0Emax=Ea=E0+15/4Emin=Eb=E0－15/4评分标准：本题每小题4分计算对Md=30Nm得４分.写出Wd=Wr但结果算错扣１分计算对Wma得４分.解题方法对或画能量示意图方法对仅计算结果算错扣１分写对JF的表达式且计算结果全对得4分，写对JF的表达式且计算结果错扣１分。7.Thefig.shownisaplatecamwithatranslatingoffsetrollerfollower.Thecamcontourconsistsoffourarcs.Usingagraphicalmethod,indicate:(20%)1)Thepitchcurveofthecamand2)TheradiusofprimecirclerP,3)Thecamangleforriseδ0,thecamangleforouterdwellδS,thecamangleforreturnδ0',thecamangleforinnerdwellδS'，4)Thelifthofthefollower,5)Findthemaximumpressureangleofthecammechanismmaxintherise.评分标准：每小题全对４分画对理论廓线得４分，错不得分.画对基圆得４分，错不得分标对０，s，0’，s’每一项得1分，错不得分。画对升程h得４分，错不得分图示对max发生位置和大小得4分，位置对但大小错扣1分